ADVANCED PROBLEMS AND SOLUTIONS Edited by Raymond E. Whitney

Transcription

1 Edited by Raymond E. Whitney Please send all communications concerning to RAYMOND E. WHITNEY, MATHEMATICS DEPARTMENT, LOCK HAVEN UNIVERSITY, LOCK HA VEN, PA This department especially welcomes problems believed to be new or extending old results. Proposers should submit solutions or other information that will assist the editor. To facilitate their consideration, all solutions should be submitted on separate signed sheets within two months after publication of the problems. PROBLEMS PROPOSES IN THIS ISSUE H-559 Proposed by N. Gauthier, Royal Military College of Canada Let n and q be nonnegative integers and show that: n a. S (gr):=g 2cos(2^/») + (-l)«+ % g = (-\y + nl qn 5 F 2qFqn " -rrf-r- " odd - nl qn - n even. ^Iq^lq^qn L n and F are Lucas and Fibonacci numbers. H-560.Proposed by H.-J. Seiffert, Berlin, Germany Define the sequences of Fibonacci and Lucas polynomials by and F 0 (x) = 0, F l (x) = l, and F n+l (x) = xf n (x) + F^x), L Q (x) = 2, L x (x) = x y and L n+l (x) = xl n (x) + L n _ x {x\ n GN, nen, respectively. Show that, for all complex numbers x and all positive integers «, Z ^ r r r I V F *W = F2 n (x)h-x)"f n (x) k=qn K\ j and [nil] I^Zl("it f ^ = L 2n (x) + (-xtl (x)- 2000]

2 SOLUTIONS Continuing... H-543 Proposed by David M Bloom, Brooklyn College of CUNY, Brooklyn, NY (Vol 36, no. 4, August 998) Find all positive nonsquare integers d such that, in the continued fraction expansion ^d = [n;a h...,a r _ h 2n], we have a x = = a r _ x =. (This includes the case r = in which there are no a's.) Solution by Charles K Cook, University of South Carolina Sumter, Sumter, SC For the case [n; 2n ], it is known (see [], p. 80) that x = [2n] satisfies x 2 = 2nx +. Thus, x - n + «Jn 2 + and so 4d = n + - which simplifies to d - n 2 +. Setting y equal to the periodic expansion and recovering a relationship for y using the usual formal manipulations on the continued fraction representation l yields the following equations iory: 0;V2S] + - 2n + ȳ 2/iy 2-2/iy-l = 0 (2n + T)y 2-4ny-2 = 0 0;Ul2^] 0 ; U U > ] (4n + l)y 2-6n-3 (6n + 2)y 2 -\0ny-5 = 0 = 0 0;l,l,l,l,l,2w] (0w + 3).y 2-6w>>-8 = 0 and, in general, if F m is the rrfi Fibonacci number, then [0; m-ones, 2n\ and y satisfies ~ Fm \ - > which can be shown by a routine inductive argument. Thus, n2l(2n-l)f m +F m+l must be integral. So both are integral. m+l (2 -l)f m ^ (2»-l)F m [ m+l m+l 92 [FEB.

3 However, gcd(f m, F m+l ). =, so 2w = (mod F m+ ). Hence, F m+l must b,e odd. Therefore, gcd(2, F m+l ) =, and the linear congruence In = (modf w+ ) always has a solution. Thus, if m is the number of ones in the continued fraction expansion, it follows that, 2, (2n-l)F m provided F m+l is odd. A few solutions are shown in the table below. m F m n = n(k), k > k k 3 f c - l 5k-2 3*-6 2*-0 n values,2,3,...,2,3,... 2,5,8,... 3,8,3,... 7,20,33,... H 32,53,... d = d(k) * 2 + l k 2 +2k k 2 +k + y 2 9k 2-2k 25k 2 -Uk+2 2 +(0& + 3)/8 69Jfc 2-40& * 2-394fc+88 * 2 +(42& + 3)/34 d values 2,5,0,... 3,8,5,... 7,32,75,... 3,74,85,... 58,425,30,... 35,064,2875,... Reference. C. D. Olds. Continued Fractions. Washington, D.C.: The Mathematical Association of America, 963. Also solved by P. Bruckman, A. Tuyl, and the proposer. Primes and FPP f s B-544 Proposed by Paul S. Bruckman, Berkeley», CA (Vol 36, mo. 4 9 August 998) Given a prime p > 5 such that Z(p) = p + l, suppose that q = \{p 2-3) and r = p 2 - p - are primes with Z(q) = q +, Z(r) = - (r -). Prove that n = pgr is a FPP (see previous proposals for definitions of the Z-function and of FPP's). Solution by the proposer For all natural m such that gcd(w, 0) =, let s m denote the Jacobi symbol (5/m), and m f = m- s m. If s is any prime * 2,5, it is well known that Z{s) \ s f. We then see that s p -s q --\, s r = g n = s p s q 8 r = +. Thus,? = ±3, q & ±3, r = ±, n = ± (mod 0). Now, if J is any prime * 2,5 and a(s) = W Z(s), then a(s) and ^-(J -) have the same parity (see this journal, Problem H-494, Vol. 33, no., Feb. 995; solution in Vol. 34, no. 2, Aug. 996, pp. 90-9). Since a(p) = a{q) =, a(r) = 2, it follows that p = q = 3, r = n = l (mod 4). Also, 32 _ 3 _ = 5? which shows that r cannot be prime if p = 3 (mod 20). Therefore, p = 7 (mod 20); this in turn implies that # = 3, r = w s (mod 20). Next, we see that Z(?) = ^ ), Z(r) = (p + l)(p-2). Then Z(n) = lcm{z(/>), Z(q\ Z(r)} = \(p 2 - \){p - 2). In order to show that n is a FPP, it suffices to show that n - = n f = 0 (mod Z(w)). Now 2000] 93

4 pq-r = ±(p 3-2p 2 -p + 2) = ±(p 2 -i)(j>-2) = Z(ny, hence, pq = r (mod Z(n)). Then/i = r 2 (mod Z(n)). Next? r + l-p(p-t), r-l = (p + l)(p-2), whence r 2 - = p(p 2 - l)(p - 2) = 2pZ(n) = 0 (mod Z(n)). Thus, n' = n-l = r 2 - = 0 (mod Z(w)), which shows that n is a FPP. Q.E.D. Note: The smallest FPP satisfying the above conditions is (p = 7). y4&0 solved by If.-/. Seiffert An Interesting Equation H-553 Proposed by Paul & Bruckman, Berkeley, CA (Vol 37, no. 3, August 999) The following Diophantine equation has the trivial solution (A, B, C, D) - (A, A, A, 0): A 3 + B 3 + C 3-3 ABC = D k, where k is a positive integer. () Find nontrivial solutions of (), i.e., with all quantities positive integers. Solution () by the proposer Let As we may easily verify: 0=exp(fwr), (2) K(a, b, c) = a 3 +A 3 + c 3-3a*c. (3) where K(a, b, c) = $(a, b, c) s(a, b0 9 c0 2 ) s(a, 0 2, cff), (4) s(a,z>,c) = a + Z> + c. (5) Given U, V, W positive integers, where at least two of them are distinct, let x = (s(u, v, W))\Y=(s(u, ve 9 we 2 ))\ z = (s(u, vd 2, we)) k. (6) From (4), it follows that Now define the following quantities: Again using (4), we see that X Z = (K(U,V,W)) k. (7) A = ±s(x, Y, Z), B = ±s(x, Y0 2, Z0), C = }s(x, Yd, Z0 2 ). (8) We now employ the following well-known expression: 27ABC = K(X,Y,Z). (9) w+^+^lj fell: o ) By trinomial expansion of the quantities defined in (8), implementing (6) and (0), we obtain the following expressions: A = F 0 (U,V,W), B = F l (U,V,W), C = F 2 (U,V,W), () where 94

5 Fj(U,V,W)= X { f+g+h=k V > 5 > ' V g-h&j (mod 3) and I * ^ is a trinomial coefficient = ' k ^U'VW, j = 0,,2, (2) From (2), it is clear that A, B, and C are positive integers. We may also easily verify the following inverse relations: Again using (4), this implies From (7) and (4), it follows that X = s(a, B,Q,Y = s(a, B0, C0 2 \ Z = s(a, B0 2, CO). (3) Thus, by reference to (), we see that we may set XYZ = K(A,B,Q. (4) K{A,B 9 Q = (K(U,V,W)) k. (5) D = K(U,V,W). (6) Accordingly, solutions (A,B,C,D) of () are given by () and (6); alternatively, A, B, and C may be obtained indirectly from (8) and (6). Note that the restriction that U, V, and W be not all identical ensures that Y and Z are positive, as of course is X. Then, from (7) and (6), it follows that D> 0, which avoids trivial solutions. Solution (2) by John Jaroma andrajih Rahntan, Gettysburg College, Gettysburg, PA After a brief historical background, we will show that, in fact, there are an infinite number of solutions of (), subject to (2): A 3 + B 3 +C 3-3ABC = D k ; () A$,C,De{l,2,...} and Jfce{2,3,...}. (2) First, in terms of a historical perspective, it appears that Diophantine equations involving cubic terms have generated considerable interest. For example, in 847, J. J. Sylvester provided sufficient conditions for the insolubility in integers of the equation Ax 3 + By 3 + Cz 3 =:Dxyz. (3) Moreover, Sylvester was able to prove that whenever (3) is insoluble, there must exist an entire family of related equations equally insoluble. His motivation for studying such equations was to break ground in the area of third-degree equations. Ultimately, Sylvester had hoped to open a new field in connection with Fermat's Last Theorem. Today, cubic equations continue to command a great deal of attention. For instance, although we know that every number (with the possible exception of those in the form 9«±4) can be expressed as the sum of four cubes, it is still not known whether every number can be expressed as the sum of four cubes with two of the cubes equal. Stated algebraically, we would like to know, if given any k, do integral solutions exist for the Diophantine equation, C 3 = Jfc. (4) (k - 76 is the first of many values of* for which an integral solution is not known.) 2000] 95

6 Perhaps an even more difficult problem exists in the question whether numbers not of the form 9n±4 can be expressed as the sum of three cubes; that is, does the equation A 3 + B 3 + C 3 = k (5) have a solution in integers \/k *9n±4 f? The first known value of k for which the problem becomes open is k Furthermore, even if we restrict ourselves to the specific case k = 3, we do not know whether ( 5,) and (4,4, - 5) are the only two solutions of (5). It is likely that Diophantine equations will continue to be an area of research for some time to come, for we know that, given an arbitrary Diophantine equation, there cannot exist an algorithm which in a finite number of steps will decide its solvability. Hilbert's Tenth Problem was demonstrated to be unsolvable by Yuri Matiyasevich in 970. Consider the following infinite sets: (I) pe{l,2,...}, * = 3/> + l, % ^ G {, 2?... } : ^ / ^ G { 2, 3,... }, D = + n\ + (n l It^f -30?! In^n^ A = D p, B = (n l /n 2 )A, C = n l A = (n 2 B). (II) * = 2, w eft 2,...}, B = D = 9n 2, A = D-n, C = D+n. Remark: We have ignored the case where p = 0, for this would imply that k = and it would then be trivial to produce infinitely many solutions of (). Proposition: Sets (I) and (II) represent disjoint families of solutions of () satisfying (2). Proof: We first prove that (I) and (II) are disjoint families of solutions of (). Since elements of (I) and (II) are ordered 4-tuples of the form (A, B, C, D) and p e {,2,...}, it follows immediately that (I) and (II) are disjoint as 3p + ^ 2. Now, to show that (I) represents an infinite set of solutions of (), we let n-r^. Hence, n x = bn for some b e{2,3,...} and D = l + b 3 +h 3 n 3-3b\ B = ba, C = nba = nb. (6) Substituting (6) into (), we get Rewriting (7), we obtain D 3p + b 3 D 3p +n 3 b 3 D 3p -3nb 2 D 3p = D 3p+ \ (7) D 3p {\ + b 3 +n 3 b 3-3nb 2 ) = D 3p+l. (8) Thus, (8) is true if and only if +b 3 +h 3 n 3-3b 2 n = D. By (6), the result follows immediately. Finally to show that (II) is also an infinite family of solutions of (), we infer from (II) that B = D = n + A and C = In + A. Substituting these quantities and the hypothesis that k - 2 into (), we obtain A 3 + (n + A) 3 +(2n + A) 3-3A(n + A)(2n + A) = (n + A) 2. (9) Simplifying (9), we obtain 9n 2 (n + A) = (n + A) 2. It now follows that (II) is a set of solutions of () if and only if 9n 2 - n - A = 0. But, by hypothesis, A-D-n- 9n 2 -n, and this produces the desired result. Also solved by B. Beasley, C Cook, andh.-j. Seiffert 96